How should I practice Request Retry Count?

Start by reading the full statement, solve the visible examples, then compare the pattern against related IBM online assessment problems in FastPrep.

Is Request Retry Count associated with IBM?

FastPrep tags this as a IBM online assessment coding practice problem when the problem catalog has that company and stage context.

Problem · Array

● Easy

IBMNEW GRADOA

An integer gap defines the maximum allowed time difference, in seconds, to consider a retry.

Two arrays are provided:

requestIds, where each element represents the request ID of a log
timestamps, where each element represents the time of the corresponding log, sorted in non-decreasing order

A retry occurs when two consecutive logs for the same request ID have a time difference of at most gap.

Compute the total number of retries across all request IDs and return the result.

getRetryCount(gap: int, requestIds: String[], timestamps: int[]) → int

gap = 10requestIds = ["r1", "r1", "r1", "r2", "r2"]timestamps = [100, 105, 200, 300, 302]return = 2

The table below shows the total number of retries for each request ID:

Request Retry Table
Request ID	Timestamps	Retry Pairs	Retry Count
`r1`	`[100, 105, 200]`	`(100, 105)`	`1`
`r2`	`[300, 302]`	`(300, 302)`	`1`

The total number of retries = 1 + 1 = 2.

Hence, the answer is 2.